per unit system practice problem solved transformers

Per Unit System – Practice Problem

Solved For Easy Understanding

Let’s understand the concept of per unit system by solving an example. In the

one-line diagram below, the impedance of various components in a power

system, typically derived from their nameplates, are presented. The task now is

to normalize these values using a common base.

Figure 1: Oneline Diagram Of A Power System

Now that you have carefully examined the system and its parameters, the

equivalent impedance diagram for the above system would look something like

the following.

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Figure 2: Impedance Diagram Of A Power System

Resistive impedance for most components have been ignored. Rotating machines

have been replaced with a voltage source behind their internal reactance.

Capacitive effects between lines and to ground are ignored as well.

To obtain the new normalized per unit impedances, first we need to figure out

the base values (Sbase, Vbase, Zbase) in the power system. Following steps will

lead you through the process.

Step 1: Assume a system base

Assume a system wide of 100MVA. This is a random assumption and chosen

to make calculations easy when calculating the per unit impedances.

So, = 100MVA

Step 2: Identify the voltage base

Voltage base in the system is determined by the transformer. For example, with

a 22/220kV voltage rating of T1 transformer, the on the primary side of T1 is

22kV while the secondary side is 220kV. It does not matter what the voltage

rating of the other components are that are encompassed by the zone.

See figure below for the voltage bases in the system.

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Figure 3: Voltage Base In The Power System

Step 3: Calculate the base impedance

The base impedance is calculated using the following formula:

Ohms…………………………………………………………………..(1)

For T-Line 1: = 484 Ohms

For T-Line 2: = 121 Ohms

For 3-phase load: = 1.21 Ohms

Step 4: Calculate the per unit impedance

The per unit impedance is calculated using the following formulas:

……………………………………………………………………………..(2)

……………………………….(3)

The voltage ratio in equation (3) is not equivalent to transformers voltage ratio.

It is the ratio of the transformer’s voltage rating on the primary or secondary

side to the system nominal voltage on the same side.

For T-line 1 using equation (2): = 0.1 pu

For T-line 2 using equation (2): = 0.5 pu

For 3-Phase load:

Power Factor:

Thus,



= 1.1495+j1.53267 Ohms

Per unit impedance of 3-phase load using equation (2)= =

0.95+j1.2667 pu

For generator, the new per unit reactance using equation (3)

= 0.2 pu

For transformer T1: = 0.2 pu




For Motor, = 0.25 pu

The equivalent impedance network with all the impedances normalized to a

common system base and the appropriate voltage base is provided below.

Per Unit Impedance Diagram

Summary:

1. Assume a Sbase for the entire system.

2. The Vbase is defined by the transformer and any off-nominal tap setting it

may have.

3. Zbase is derived from the Sbase and Vbase.

4. The new per unit impedance is obtained by converting the old per unit

impedance on old base values to new ones. See equations (2) and (3).

* * * * *



base values per unit per unit impedances per unit system per unit value Power

transformers

32 Responses to Per Unit System – Practice Problem Solved

For Easy Understanding

IJAJIJAJIJAJIJAJ says:

what do you mean by….. s*-??

anananan says:

Can we find the short circuit current at each end?

samuelsamuelsamuelsamuel says:

A load of 50mw at 0.8 power factor lagging is taken from the 33kv.

( taking a base MVA of 100mva), calculate the terminal voltage of

the synchronous machine? (Please help me solve this question)

thanks

NikhilNikhilNikhilNikhil says:

very useful thanks

alshaiaalshaiaalshaiaalshaia says:

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How can we determine the voltage on the bus 1

BRianBRianBRianBRian says:

Do you know how to find the voltage at the bus?

Thanks

tahseentahseentahseentahseen says:

Hi

how we can find the voltage in bus1 in PU and in volte

noanoanoanoa says:

tanks alot save me alot of stress

abiabiabiabi says:

. Obtain the per unit impedance(reactance) diagram of the power

system shown in the fig

G1 : 30MVA , 10.5KV, X?=1.6 ?

G2 : 15MVA , 6.6KV, X?=1.2 ?

G3 : 25MVA , 6.6KV, X?=0.56 ?

T1 (3 phase): 15MVA , 33/11KV , X= 15.2 ? per phase on the high

tension side

T2 (3 phase): 15MVA , 33/6.2KV , X= 16 ? per phase on the high

tension side

Transmission line : 20.5 ohm per phase

Load A : 15MW , 11KV , 0.9 p.f lagging

Load B : 40MW , 6.6KV , 0.85 p.f lagging

5

abiabiabiabi says:

how to convert ohms value to per unit value

Lee TaylorLee TaylorLee TaylorLee Taylor says:

Hi, great article – thanks very much! I have a similar problem to



solve but I am struggling with the Zact calculation. My inputs are

Vrated = 4.16kV, S = 2MVA <-36.87. Can you help?!

bhanubhanubhanubhanu says:

awesome

kaushik vastarparakaushik vastarparakaushik vastarparakaushik vastarpara says:

its really bcoz by reading this my confusion abut selection of base

nd other is very clear…sommust read it frend /…thank u

AdminAdminAdminAdmin says:

@Pavan @Mike: That’s a typo. Correct values are now shown in

the calculations. Since the ratio of Vbase_old/Vbase_new is the

same, the end result, therefore, does not change. Appreciate the

feedback.

PavanPavanPavanPavan says:

This is really helpful. I didn’t really got it when reading through

this, but when I saw the below comment by Mike, it seems like a

question worth answering. However the content is really clear and

understandable. Keep up the good work!

Thanks,

mikemikemikemike says:

I don’t understand one part: When calculating Xtl2 you are using

(22/22) which is reflected from where? Vbase in T2 is 220 primary

and 11 secondary, so where does 22 come from?

The same for Xtl4.

markmarkmarkmark says:

will the impedance or p.u impedance in each line will be like in

series? will the current for the PRIMARY AND SECONDARY of the

transformer now be equal??? how will i find the actual line current

for each line and for the whole system…



chrischrischrischris says:

A single phase ,350 kva, 1380v generator has an internal

impedance Zg of j6 ohms. The generator is used to supply a load of

250kva/440v at power factor 0.78 lagging. determine: the turns

ratio of the transfomer, the impedance per km if the line between

the generator and the transformer is 5km, the voltage regulation

of the system.

Using the ratings of the generator as base values determine the

generated per unit voltage that is required to produce a full load

current under short circuit condition.

CAN SOMEONE HELP ME WIT THESE CALCULATION PLZ!!!


Kam,

Once you have the impedance network, use the current division

rule to determine the current flowing each line. I am not sure I

understand “voltage at 3″, if bus 3 is faulted (3ph) then it is zero

otherwise it should be the same as nominal voltage as seen on the

secondary side of the transformer.

I will solve one for the currents in the future but for now, you will

have to learn how to reduce a circuit (using KVL and KCL) to

determine the currents.

kamkamkamkam says:

sorry but i didnt get my reply yet, so could you pls help me

out???????? thanks a lot

karthikkarthikkarthikkarthik says:

very well explained but could you pls show me how to calculate

voltage and current in both lines, will be very greatful , thanks a

lot……….

kamkamkamkam says:

It is really well explained but could you pls show me how to

calculate voltage at but 3 and current in both lines, will be very

greatful , thanks a lot



manishmanishmanishmanish says:

what if transformers are connected in star and delta connection?

AnayatAnayatAnayatAnayat says:

i am very new to Power side , so i really dont know abt all these

concepts , what we only have T1 and T2 , and all the rating given

are three phase line to line ? how we ll solve it then?

richaricharicharicha says:

very nicely explained….to the point and complete..thanks a lot

SanketSanketSanketSanket says:

VERY NICELY EXPLAINED THANK-YOU ………

I WILL VISIT WEBSITE AGAIN FOR FURTHER REFERENCES.

BABULS RAJBABULS RAJBABULS RAJBABULS RAJ says:

Thank u so much…..after searching for a proper explanation for

the same in so many sites, i got it finally from your site. Clear

explanation with proper diagrams with multi colour…….very nice

…..


Nice catch. Fixed it. Thanks.

Renjith MRenjith MRenjith MRenjith M says:

Commendable work. But there is a small error. The per-unit

system is the ratio of two quantities of the same units. Therefore

it is unitless. Well that is what I know. So accordingly we specify

the per-unit quatities as just ‘P.U’. So you need to remove the

‘Ohms’ from the text and insert ‘p.u’



AlfredoAlfredoAlfredoAlfredo says:

It was very useful, but it is short because is necessary to get the

complete solution, any way I liked.

Abdul RauffAbdul RauffAbdul RauffAbdul Rauff says:

Very Good Info About PSA.Thanks Alot

HilaryHilaryHilaryHilary says:

Protection engieering, i have been give the reactance as Xd’ to

calculate faults on a system do i convert to Xd” how do i do this

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